Best Time to Buy and Sell Stock III

Best Time to Buy and Sell Stock III
首先想得的思路便是遍历
遍历数组，将数组分成左右2部分，分布对左右部分进行一次买卖，profit = leftProfit + rightProfit。

public class Solution {        public int maxProfit(int[] prices) {        if(prices.length == 0){return 0;}        int maxProfit = getMaxProfitByOneStep(prices);        for(int i = 1;i<prices.length-1;i++)        {            int profit = getMaxProfitByOneStep(Arrays.copyOfRange(prices, 0, i+1))                    +getMaxProfitByOneStep(Arrays.copyOfRange(prices, i+1, prices.length));            if(profit > maxProfit) maxProfit = profit;        }        return maxProfit;    }    int getMaxProfitByOneStep(int[] prices)    {        int maxProfit = 0;        int n = prices.length;        int price = prices[n-1];        for(int i = n-2;i>=0;i--)        {            int profit = price - prices[i];            if(prices[i] > price)            {                price = prices[i];            }            maxProfit = maxProfit > profit ? maxProfit : profit;        }        return maxProfit;    }}

然而运行超时！
发现一些小规律。
1、如“1，2，3，4，5，3”递增，显然没必要从1到5都要split，然后求profit。直接在5处划分
2、如“2，5，4，3，2，1，4”递减，直接在1处划分，因为{“2,5,4,3,2”，“1，4”}={“2，5，4，3”，“2，1，4”}={“2，5，4”，“3，2，1，4”}={“2，5”，“4，3，2，1，4”}

public class Solution {        public int maxProfit(int[] prices) {        if(prices.length == 0){return 0;}        int maxProfit = getMaxProfitByOneStep(prices);        int right_begin = 0;        int left_end = 0;        for(int i = 1;i<prices.length-1;i++)        {            while(i<prices.length-2&&prices[i+1]>=prices[i]){i++;left_end = i;}            while(i<prices.length-2&&prices[i+1]<prices[i]){i++;right_begin = i; }            if(left_end<i){left_end = i;}            if(right_begin<i){right_begin=i+1;}            int profit = getMaxProfitByOneStep(Arrays.copyOfRange(prices, 0, left_end+1))                    +getMaxProfitByOneStep(Arrays.copyOfRange(prices, right_begin, prices.length));            if(profit > maxProfit) maxProfit = profit;        }        return maxProfit;    }    int getMaxProfitByOneStep(int[] prices)    {        int maxProfit = 0;        int n = prices.length;        int price = prices[n-1];        for(int i = n-2;i>=0;i--)        {            int profit = price - prices[i];            if(prices[i] > price)            {                price = prices[i];            }            maxProfit = maxProfit > profit ? maxProfit : profit;        }        return maxProfit;    }}

接下来自然而然想到动态规划！！
maxProfit = max{leftProfit[i]+rightProfit[j]};

public class Solution {    public int maxProfit(int[] prices) {        int n = prices.length;        int[] leftProfit = new int[n];        int[] rightProfit = new int[n];        int min = Integer.MAX_VALUE;        int maxProfit = 0;        for(int i = 0;i<n;i++)        {            min = Math.min(min, prices[i]);            maxProfit = Math.max(maxProfit, prices[i]-min);            leftProfit[i] = maxProfit;        }        int max = Integer.MIN_VALUE;        maxProfit = 0;        for(int i = n-1;i>=0;i--)        {            max = Math.max(max, prices[i]);            maxProfit = Math.max(maxProfit, max - prices[i]);            rightProfit[i] = maxProfit;        }        maxProfit = 0;        for(int i = 0 ;i<n;i++)        {            maxProfit = Math.max(maxProfit, leftProfit[i]+rightProfit[i]);        }        return maxProfit;    }}

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